91Ó°ÊÓ

We are given that,

The maxwell speed distribution can then be written

$D\left(v\right)=\frac{4}{\sqrt{\mathrm{Ï€}}}\frac{v^2}{{v_o}^2}t\frac{-3}{2}e{\raisebox{1ex}{$-v^2$}\!\left/ \!\raisebox{-1ex}{$v_0$}\right.}^{2}t$

$T=300k,600k$

Let us define the constant $v_0≡\sqrt{2kT/m}$for $T=300k$. The mass of a nitrogen molecule is$28u$,

So, $v_o=\sqrt{\frac{2k\left(300k\right)}{m}}=\sqrt{\frac{2(1.38×{10}^{-23}J/k)\left(300k\right)}{28(1.66×{10}^{-27}kg)}}=422m/s$

The maxwell speed distribution can then be written

$D\left(v\right)=\frac{4}{\sqrt{\mathrm{Ï€}}}\frac{v^2}{{v_o}^2}t\frac{-3}{2}e{\raisebox{1ex}{$-v^2$}\!\left/ \!\raisebox{-1ex}{$v_0$}\right.}^{2}t$,

Where t is the temperature in units of $300k$. To plot this function for$t=1$and $t=2,$

I gave $Mathematical$the following instruction:

$$\begin{gathered} v_0=422; \\ maxwell[t,v]:=2.257×(v^2/{v^3}_0)×t^{-(-1.5)}×Exp[-v^2/({v_0}^2×t\left)\right] \\ Plot\left[\right\{maxwell[1,v],maxwell[2,v]\},\{v,0,1700\} \end{gathered}$$

Here the plot graph

Notice that the area under each curve is equal to 1. Therefore, as the location of the peak

Moves to the right (in proportion to $\sqrt{T}$), its height must decrease.